Rotations, Complex Matrices and Real Matrices

In summary, Tapp's Section 1 in Chapter 2 discusses the function \rho_n : M_n ( \mathbb{C} ) \rightarrow M_{2n} ( \mathbb{R} ), which relates complex matrices to real matrices. This function is defined as R_{\rho_n(A)} : \Bbb R^{2n} \to \Bbb R^{2n}, where R_A(X) = XA and R_{\rho_n(A)}(Y) = Y\rho_n(A). The relationship between \rho_n and f_n is given by R_{\rho_n}(A)(f_n(c_1 + \mathbf{i}d_1,\ldots,
  • #1
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I am reading Kristopher Tapp's book: Matrix Groups for Undergraduates.

I am currently focussed on and studying Section 1 in Chapter2, namely:

"1. Complex Matrices as Real Matrices".I need help in fully understanding what Tapp is saying in this section regarding the function

\(\displaystyle \rho_n \ : \ M_n ( \mathbb{C} ) \rightarrow M_{2n} ( \mathbb{R} ) \)Section 1, Chapter2 reads as follows:
View attachment 3990
View attachment 3991I am having trouble fully understanding how the function:

\(\displaystyle \rho_n \ : \ M_n ( \mathbb{C} ) \rightarrow M_{2n} ( \mathbb{R} \)

relates to \(\displaystyle f_n\) and \(\displaystyle R_{ \rho_n (A) }\) ...

For example if

\(\displaystyle \rho_2 \begin{pmatrix} a+bi & c+di \\ e + fi & h+ji \end{pmatrix}
= \begin{pmatrix} a & b & c & d \\ -b & a & -d & c \\ e & f & h & j \\ -f & e & -j & h \end{pmatrix}\)
My question is what is

\(\displaystyle f_n\) and \(\displaystyle R_{ \rho_n (A) }\)

in this case, and how exactly do these expressions relate to \(\displaystyle \rho_n\) ... ...?
Hope someone can help?
Now in the above the linear transformation \(\displaystyle R_A\) is mentioned ... ... \(\displaystyle R_A\) is defined in Tapp Ch. 1, Section 5 as follows:
https://www.physicsforums.com/attachments/3992
View attachment 3993
 
Last edited:
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  • #2
Hi Peter,

I don't know if I have understood your issue.

For example, when $n=2$

$f_{2}:\Bbb{C}^2\longrightarrow \Bbb{R}^4$ is given by

$f_{2}(a+bi,c+di)=(a,b,c,d)$

And $R_{\rho_{2}(A)}$ is the multiplication of a row vector times this $4\times 4$ matrix you have typed on.

The relation $\rho_{n}$ gives you is how you can define the same application modulo the canonical isomorphism from $\Bbb{C}^{n}$ to $\Bbb{R}^{n}$ , looking at it like a complex function or a real one.
 
  • #3
Peter said:
I am reading Kristopher Tapp's book: Matrix Groups for Undergraduates.

I am currently focussed on and studying Section 1 in Chapter2, namely:

"1. Complex Matrices as Real Matrices".I need help in fully understanding what Tapp is saying in this section regarding the function

\(\displaystyle \rho_n \ : \ M_n ( \mathbb{C} ) \rightarrow M_{2n} ( \mathbb{R} ) \)Section 1, Chapter2 reads as follows:

I am having trouble fully understanding how the function:

\(\displaystyle \rho_n \ : \ M_n ( \mathbb{C} ) \rightarrow M_{2n} ( \mathbb{R} \)

relates to \(\displaystyle f_n\) and \(\displaystyle R_{ \rho_n (A) }\) ...

For example if

\(\displaystyle \rho_2 \begin{pmatrix} a+bi & c+di \\ e + fi & h+ji \end{pmatrix}
= \begin{pmatrix} a & b & c & d \\ -b & a & -d & c \\ e & f & h & j \\ -f & e & -j & h \end{pmatrix}\)
My question is what is

\(\displaystyle f_n\) and \(\displaystyle R_{ \rho_n (A) }\)

in this case, and how exactly do these expressions relate to \(\displaystyle \rho_n\) ... ...?
Hope someone can help?
Now in the above the linear transformation \(\displaystyle R_A\) is mentioned ... ... \(\displaystyle R_A\) is defined in Tapp Ch. 1, Section 5 as follows:

Peter,

First, let me say that there is something confusing going on in the first section you posted. As written, the matrix $B$ represents clockwise rotation through angle $\theta$, not counterclockwise rotation. It can't be a difference in convention because the computation $R_B(r\cos \phi, r\sin \phi) = (r\cos(\theta + \phi), r\sin(\theta + \phi))$ suggests that the intended matrix is

$$B = \begin{pmatrix}\cos \theta & -\sin \theta\\\sin \theta & \cos \theta\end{pmatrix}$$

However, after the computation of $R_A(e^{i\theta})$, the clockwise convention is used! So I'll be using the clockwise convention in this discussion.

Given a matrix $A\in M_n(\Bbb C)$, $\rho_n(A)\in M_{2n}(\Bbb R)$. So $A$ represents the linear map $R_A : \Bbb C^n \to \Bbb C^n$ and $\rho_n(A)$ represents the linear map $R_{\rho_n(A)} : \Bbb R^{2n} \to \Bbb R^{2n}$. That is, $R_A(X) = XA$ for all $X\in M_n(\Bbb C)$ and $R_{\rho_n(A)}(Y) = Y\rho_n(A)$ for all $Y\in M_{2n}(\Bbb R)$. We want to relate $R_A$ and $R_{\rho_n(A)}$ by a linear map from $\Bbb C^n \to \Bbb R^{2n}$, because then we can determine $R_{\rho_n(A)}$, and hence $\rho_n(A)$. This is where $f_n$ comes in. To have $R_{\rho_n(A)} \circ f_n = f_n \circ R_A$ means that

$$R_{\rho_n}(A)(f_n(c_1 + \mathbf{i}d_1,\cdots, c_n + \mathbf{i}d_n)) = f_n(R_A(c_1 + \mathbf{i}d_1,\ldots, c_n + \mathbf{i}d_n))$$

that is, if $R_A(c_1 + \mathbf{i}d_1,\ldots, c_n + \mathbf{i}d_n) = (x_1 + \mathbf{i}y_1,\ldots, x_n + \mathbf{i}y_n)$, then

$$R_{\rho_n}(A)(c_1,d_1,\ldots,c_n,d_n) = (x_1,y_1,\ldots, x_n,y_n).$$

Unraveling this a bit further, we find that if

$$A = \begin{pmatrix}a_{11} + \mathbf{i}b_{11}&\cdots&a_{1n} + \mathbf{i}b_{1n}\\a_{21} + \mathbf{i}b_{21}&\cdots&a_{2n} + \mathbf{i}b_{nn}\\ \vdots & & \vdots\\a_{n1} + \mathbf{i}b_{n1}&\cdots &a_{nn} + \mathbf{i}b_{nn}\end{pmatrix}$$

then for $n = 1$,

$$\rho_n(A) = \begin{pmatrix}a_{11} & b_{11}\\-b_{11} & a_{11}\end{pmatrix}$$

and for $n > 1$,

$$\rho_n(A) = \begin{pmatrix}\rho_1(a_{11} + \mathbf{i}b_{11})&\cdots &\rho_1(a_{1n} + \mathbf{i}b_{1n})\\ \rho_1(a_{21} + \mathbf{i}b_{21})&\cdots&\rho_1(a_{2n} + \mathbf{i}b_{2n})\\ \vdots & & \vdots\\ \rho_1(a_{n1} + \mathbf{i}b_{n1}) & \cdots & \rho_1(a_{nn} + \mathbf{i}b_{nn})\end{pmatrix}$$

As you can see here, I have written the expression of $\rho(A)$ in block form, where each block is determined by the image of an entry in $A$ under $\rho_1$.
 
  • #4
Hi Euge,

Matrix $B$ is clockwise if you work with a column vector $v$ and then compute $Bv$, but I understand he is working with row vectors $w$, so in this case $wB$ is a counterclockwise rotation. It is not usual to work with row vectors and postmultiplicate the matrix, but it's still correct.
 
  • #5
Yes, Fallen Angel, that's right. I was going to edit that but thanks for adding that comment.
 
  • #6
Euge said:
Yes, Fallen Angel, that's right. I was going to edit that but thanks for adding that comment.
Thanks to the posts of Euge and Fallen Angel ... and some further work and reflection, I now have a much better understanding of Tapp's section on "Complex matrices as real matrices" ... ... i was really confused there for a while ... so many thanks to Euge and Fallen Angle for their help ...Just a further issue however ... in the initial case where:

\(\displaystyle \rho_1(a+bi) = \begin{pmatrix} a & b \\ -b & a \end{pmatrix}\)

it is clear from the geometry and Tapp's analysis that:

\(\displaystyle a = \cos \theta\) and \(\displaystyle b = \sin \theta\)... ... BUT ... what is the situation with respect to \(\displaystyle a,b,c, d, e, f, h, j\) in \(\displaystyle \rho_2 \begin{pmatrix} a+bi & c+di \\ e + fi & h+ji \end{pmatrix}
= \begin{pmatrix} a & b & c & d \\ -b & a & -d & c \\ e & f & h & j \\ -f & e & -j & h \end{pmatrix}\)Hope someone can help?

Peter
 
  • #7
Hi Peter,

In that case, you got the same situation BUT you are working in a 2-dimensional complex vector space, so you have 4 rotation angles, each one corresponding to the rotation of one of the two entry complex numbers with respect of one of the two complex planes , but this can't be drawn or easily visualized.

The cosinus and sinus of this angles are $(a,b), (c,d), (e,f)$ and $ (g,h)$.
 
  • #8
Fallen Angel said:
Hi Peter,

In that case, you got the same situation BUT you are working in a 2-dimensional complex vector space, so you have 4 rotation angles, each one corresponding to the rotation of one of the two entry complex numbers with respect of one of the two complex planes , but this can't be drawn or easily visualized.

The cosinus and sinus of this angles are $(a,b), (c,d), (e,f)$ and $ (g,h)$.
Thanks for your further help Fallen Angel ... I obviously need to learn more about complex vector spaces ...

Thanks again,

Peter
 

FAQ: Rotations, Complex Matrices and Real Matrices

What is a rotation matrix?

A rotation matrix is a type of complex matrix that is used to represent rotations in a three-dimensional space. It is a square matrix with a dimension of 3x3 and has a special property that it preserves the length of vectors and the angles between them during rotations.

How can I determine if a matrix is real or complex?

A real matrix is a matrix that contains only real numbers, while a complex matrix contains at least one complex number. To determine if a matrix is real or complex, you can check if all the entries in the matrix are real numbers. If at least one entry is a complex number, then the matrix is complex.

Can a rotation matrix be a real matrix?

Yes, a rotation matrix can be a real matrix. This happens when the rotation is around the z-axis, and the angle of rotation is a multiple of 90 degrees. In this case, the rotation matrix will only contain real numbers and will not involve any complex numbers.

How do I perform matrix multiplication with complex matrices?

To perform matrix multiplication with complex matrices, you need to use the rules of complex multiplication. This involves multiplying the corresponding elements and then adding them together. Also, remember to use the distributive property when multiplying complex numbers.

Can a real matrix have complex eigenvalues?

No, a real matrix cannot have complex eigenvalues. This is because the eigenvalues of a matrix are the solutions to the characteristic equation, which is a polynomial with real coefficients. Therefore, the eigenvalues must also be real numbers.

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